Question

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = 1/2 ∠A.

Answer 1

Therefore it's proved that ∠D = 1/2 ∠A.

Consider the attached diagram while going through the following steps,

from figure, it's clear that,

∠ ABE and ∠ ABC are linear pairs.

∴ ∠ ABE + ∠ ABC = 180°

2x + ∠ ABC = 180°

∠ ABC = 180° - 2x

In Δ ABC

∠ A + ∠ B + ∠ C =180°

∠ A + 180° - 2x + 2y =180°

∠ A = 2 (x-y).........(1)

In Δ DBC

∠ D + ∠ B + ∠ C =180°

∠ D + 180° - x + y =180°

∠ D = (x-y).........(2)

Comparing equations (1) and (2), we get,

∠D = 1/2 ∠A.

Hence proved.